0.91/0.91	YES
0.91/0.91	
0.91/0.91	DP problem for innermost termination.
0.91/0.91	P =
0.91/0.91	  f5#(x1, x2) -> f4#(x1, x2) 
0.91/0.91	  f4#(I0, I1) -> f3#(I0, 0) 
0.91/0.91	  f3#(I2, I3) -> f1#(rnd1, 2 + I3) [rnd1 = 2 + 2 + I3]
0.91/0.91	  f1#(I4, I5) -> f3#(I4, I5) [1 + I5 <= 20]
0.91/0.91	R = 
0.91/0.91	  f5(x1, x2) -> f4(x1, x2) 
0.91/0.91	  f4(I0, I1) -> f3(I0, 0) 
0.91/0.91	  f3(I2, I3) -> f1(rnd1, 2 + I3) [rnd1 = 2 + 2 + I3]
0.91/0.91	  f1(I4, I5) -> f3(I4, I5) [1 + I5 <= 20]
0.91/0.91	  f1(I6, I7) -> f2(I6, I7) [20 <= I7]
0.91/0.91	
0.91/0.91	The dependency graph for this problem is:
0.91/0.91	  0 -> 1
0.91/0.91	  1 -> 2
0.91/0.91	  2 -> 3
0.91/0.91	  3 -> 2
0.91/0.91	Where:
0.91/0.91	  0) f5#(x1, x2) -> f4#(x1, x2) 
0.91/0.91	  1) f4#(I0, I1) -> f3#(I0, 0) 
0.91/0.91	  2) f3#(I2, I3) -> f1#(rnd1, 2 + I3) [rnd1 = 2 + 2 + I3]
0.91/0.91	  3) f1#(I4, I5) -> f3#(I4, I5) [1 + I5 <= 20]
0.91/0.91	
0.91/0.91	We have the following SCCs.
0.91/0.91	  { 2, 3 }
0.91/0.91	
0.91/0.91	DP problem for innermost termination.
0.91/0.91	P =
0.91/0.91	  f3#(I2, I3) -> f1#(rnd1, 2 + I3) [rnd1 = 2 + 2 + I3]
0.91/0.91	  f1#(I4, I5) -> f3#(I4, I5) [1 + I5 <= 20]
0.91/0.91	R = 
0.91/0.91	  f5(x1, x2) -> f4(x1, x2) 
0.91/0.91	  f4(I0, I1) -> f3(I0, 0) 
0.91/0.91	  f3(I2, I3) -> f1(rnd1, 2 + I3) [rnd1 = 2 + 2 + I3]
0.91/0.91	  f1(I4, I5) -> f3(I4, I5) [1 + I5 <= 20]
0.91/0.91	  f1(I6, I7) -> f2(I6, I7) [20 <= I7]
0.91/0.91	
0.91/0.91	We use the reverse value criterion with the projection function NU:
0.91/0.91	  NU[f1#(z1,z2)] = 20 + -1 * (1 + z2)
0.91/0.91	  NU[f3#(z1,z2)] = 20 + -1 * (1 + (2 + z2))
0.91/0.91	
0.91/0.91	This gives the following inequalities:
0.91/0.91	  rnd1 = 2 + 2 + I3  ==>  20 + -1 * (1 + (2 + I3)) >= 20 + -1 * (1 + (2 + I3))
0.91/0.91	  1 + I5 <= 20  ==>  20 + -1 * (1 + I5) > 20 + -1 * (1 + (2 + I5))  with  20 + -1 * (1 + I5) >= 0
0.91/0.91	
0.91/0.91	We remove all the strictly oriented dependency pairs.
0.91/0.91	
0.91/0.91	DP problem for innermost termination.
0.91/0.91	P =
0.91/0.91	  f3#(I2, I3) -> f1#(rnd1, 2 + I3) [rnd1 = 2 + 2 + I3]
0.91/0.91	R = 
0.91/0.91	  f5(x1, x2) -> f4(x1, x2) 
0.91/0.91	  f4(I0, I1) -> f3(I0, 0) 
0.91/0.91	  f3(I2, I3) -> f1(rnd1, 2 + I3) [rnd1 = 2 + 2 + I3]
0.91/0.91	  f1(I4, I5) -> f3(I4, I5) [1 + I5 <= 20]
0.91/0.91	  f1(I6, I7) -> f2(I6, I7) [20 <= I7]
0.91/0.91	
0.91/0.91	The dependency graph for this problem is:
0.91/0.91	  2 -> 
0.91/0.91	Where:
0.91/0.91	  2) f3#(I2, I3) -> f1#(rnd1, 2 + I3) [rnd1 = 2 + 2 + I3]
0.91/0.91	
0.91/0.91	We have the following SCCs.
0.91/0.91	  
0.91/3.89	EOF